How to show $I_p(a,b) = \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$ 
Show that $$I_p(a,b) = \frac{1}{B(a,b)}\int_0^p u^{a-1}(1-u)^{b-1}~du\\= \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$$ when $a,b$ are positive integers.

I have no idea how to proceed. Please help.
 A: Hint: Use $\frac{1}{B(a,b)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}=\frac{(a+b-1)!}{(a-1)!(b-1)!}$ and integration by parts to evaluate the integral:
$$I=\int_{0}^{1}u^{a-1}(1-u)^{b-1} = \left.\frac{1}{a}u^a(1-u)^{b-1}\right|_{0}^{p}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I= \frac{1}{a}p^a(1-p)^{b-1}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I = \frac{1}{a}p^a(1-p)^{b-1} + \frac{b-1}{a(a+1)}p^{a+1}(1-p)^{b-2} +\frac{(b-1)(b-2)}{a(a+1)}\int_{0}^{p}u^{a+2}(1-u)^{b-3}\,du = \ldots $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\,{\rm I}_{p}\pars{a,b}={1 \over \,{\rm B}\pars{a,b}}
    \int_{0}^{p}u^{a - 1}\ \pars{1 - u}^{b - 1}\,\dd u
    =\sum_{j\ =\ a}^{a + b - 1}
    {a + b - 1 \choose j}p^{j}\pars{1 - p}^{a + b - 1 -j}\,\,\,:\ {\large ?}}$.

We'll perform a 'direct evaluation' of the integral. That is accomplished with a suitable change of variable: $\ds{u\ \mapsto\ p - u}$:

\begin{align}&\int_{0}^{p}u^{a - 1}\ \pars{1 - u}^{b - 1}\,\dd u
=\int_{0}^{p}\pars{p - u}^{a - 1}\ \pars{1 - p + u}^{b - 1}\,\dd u
\\[5mm]&=\int_{0}^{p}\sum_{k\ =\ 0}^{a - 1}{a - 1 \choose k}
p^{a - 1 - k}\,\,\pars{-1}^{k}\,u^{k}\
\sum_{j\ =\ 0}^{b - 1}{b - 1 \choose j}
\pars{1 - p}^{b - 1 - j}\,\,\,\,u^{j}\,\,\,\dd u
\\[5mm]&=\sum_{k\ =\ 0}^{a - 1}\ \sum_{j\ =\ 0}^{b - 1}
{a - 1 \choose k}{b - 1 \choose j}\pars{-1}^{k}\,p^{a - 1 - k}\,\,
\pars{1 - p}^{b - 1 - j}\,\,\,{p^{k + j + 1} \over k + j + 1}
\\[5mm]&=\sum_{k\ =\ 0}^{a - 1}\ \sum_{j\ =\ 0}^{b - 1}
{a - 1 \choose k}{b - 1 \choose j}
{\pars{-1}^{k}\, \over  k + j + 1}\,p^{a + j}\,\,\pars{1 - p}^{b - 1 - j}
\\[5mm]&=\sum_{j\ =\ a}^{a + b - 1}\ \sum_{k\ =\ 0}^{a - 1}
{a - 1 \choose k}{b - 1 \choose j - a}
{\pars{-1}^{k} \over  k + j - a + 1}\,p^{j}\,\,\pars{1 - p}^{a + b - 1 - j}
\\[5mm]&=\sum_{j\ =\ a}^{a + b - 1}\bracks{{b - 1 \choose j - a}
\dsc{\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{\pars{-1}^{k} \over  k + j - a + 1}}}
p^{j}\,\,\pars{1 - p}^{a + b - 1 - j}\tag{1}
\end{align}

\begin{align}
&\dsc{\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{\pars{-1}^{k} \over  k + j - a + 1}}
=\sum_{k\ =\ 0}^{a - 1}{a - 1 \choose k}\pars{-1}^{k}
\int_{0}^{1}t^{k + j - a}\,\,\,\dd t
\\[5mm]&=\int_{0}^{1}t^{j - a}
\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}\pars{-t}^{k}\,\,\dd t
=\int_{0}^{1}t^{j - a}\,\,\pars{1 - t}^{a - 1}\,\,\dd t
=\,{\rm B}\pars{j -a + 1,a}
\\[5mm]&={\Gamma\pars{j - a + 1}\Gamma\pars{a} \over \Gamma\pars{j + 1}}
={\pars{j - a}!\,\pars{a - 1}! \over j!}\tag{2}
\end{align}

With $\pars{1}$ and $\pars{2}$:
\begin{align}&{b - 1 \choose j - a}
\dsc{\sum_{k\ =\ 0}^{a - 1} {a - 1 \choose k}{\pars{-1}^{k} \over  k + j - a + 1}}
={\pars{b - 1}! \over \pars{j - a}!\pars{b - 1 - j + a}!}\,
{\pars{j - a}!\,\pars{a - 1}! \over j!}
\\[5mm]&={\pars{a + b - 1}! \over j!\pars{a + b - 1 - j}!}\,
{\pars{a - 1}!\pars{b - 1}! \over \pars{a + b - 1}!}
={a + b - 1 \choose j}\,{\Gamma\pars{a}\Gamma\pars{b} \over \Gamma\pars{a + b}}
\\[5mm]&={a + b - 1 \choose j}\,\,{\rm B}\pars{a,b}
\end{align}


Replace this result in $\pars{1}$ to get the final result.

A: Hint: 
1) Convince yourself, from its integral representation, that
$$
I_{p}(a, b){}={}I_{p}(a, b-1){}+{}\dfrac{p^a (1-p)^{b-1}}{(b-1)B(a,b-1)}\,.
$$
2) $$
I_{p}(a, b){}={}I_{p}(a+1, b){}+{}\dfrac{p^a (1-p)^{b}}{(a)B(a,b)}\,.
$$
3) $B(a,b){}={}\dfrac{(a-1)!(b-1)!}{(a+b-1)!}\,$.
4) Show the result recursively!
A: Hint: Assume $a,b > 0$.
First, Show that 
$$
I_p(a,b)= 
\int_0^p B(a,b)^{-1} t^{a-1}(1-t)^{b-1} dt = {a+b-1 \choose a}p^a(1-p)^{b-1} + I_p(a+1,b-1)
$$
You can do this many ways and one way is by denoting
$$
f:=(1-t)^b-1 \qquad\qquad g:=t^{a}/a
$$
and apply integration by parts with the following identity 
$$ 
B(a,b)^{-1} = \frac{a}{b-1}B(a+1,b-1)^{-1}
$$
This should be straightforward to work out and just solve it recursively with the final summation term 
$$
I_p(a+b-1,1) = p^{a+b-1}
$$
and you are done.
